Solve linear equations in the rationals and Z_n

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Question statement

Solve each of the following equations, first over $\mathbb{Q}$ , the rational numbers and then over $\mathbb{Z}_{\var{n}}$ .

Solutions to the equations over $\mathbb{Q}$ must be entered as integers or fractions.

The solutions in $\mathbb{Z}_{\var{n}}$ must be in the set $\{0,1,\ldots,\var{n-1}\}$

Give any introductory information the student needs.

			Name	Type	Generated Value

a1	integer	-13
a3	number	8
n	integer	13
b4	integer	4
s	integer	1
a2	integer	2
a5	number	5
b1	integer	0
b2	integer	5
b3	integer	2
invb4	number	10
invb3	number	7

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Generated value: `integer`

-13

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Part b) Gap-fill
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$\simplify{{b2}*x-{b2*a2}=0}$

Solution in $\mathbb{Q}=$ Gap 0.

Solution in $\mathbb{Z}_{\var{n}}=$ Gap 1.

Advice

a)

The solution in $\mathbb{Q}$ is $x=\var{a2}$ .

The solution in $\mathbb{Z}_{\var{n}}$ is also $x=\var{a2}$ .

b)

The solution in $\mathbb{Q}$ is $x= \frac{\var{b3+1}}{\var{b3}}$ .

The solution in $\mathbb{Z}_{\var{n}}$ is given by $x=\var{b3+1}\times\var{b3}^{-1} \pmod{\var{n}}$ .

We find that $\var{b3}^{-1} = \var{invb3}$ in $\mathbb{Z}_{\var{n}}$ . See below for one method for finding the inverse.

Hence the solution is $x=\var{b3+1}\times\var{invb3} \equiv \var{(b3+1)*invb3}=\var{a3} \pmod {\var{ n}}$ .

Finding the inverse:

Use the Euclidean Algorithm to find $s$ and $t$ such that $\simplify{{b3}s+{n}t = gcd({b3},{n})=1}$ .

It follows that $s \bmod n$ is the inverse of $\var{b3}$ in $\mathbb{Z}_{\var{n}}$ .

In this case one such application gives:

{describegcd(n,b3)}

Using this we obtain:

$\simplify[]{{extendedgcd2(n,b3)}*{b3}+{extendedgcd1(n,b3)}*{n}=1}$

Hence the inverse of $\var{b3}$ in $\mathbb{Z}_{\var{n}}$ is $\simplify{{extendedgcd2(n,b3)}} \equiv \var{invb3} \pmod{\var{n}}$ .

c)

The solution in $\mathbb{Q}$ is $x = \simplify{{-s}*({b4+2}/{b4})}$ .

The solution in $\mathbb{Z}_{\var{n}}$ is given by $x=\var{-s*(b4+2)}\times\var{b4}^{-1} \pmod{\var{n}}$ .

We find that $\var{b4}^{-1}=\var{invb4}$ in $\mathbb{Z}_{\var{n}}$ . See below for one method for finding the inverse.

Hence the solution is $x=\var{-s*(b4+2)}\times\var{invb4}=\var{-s*(b4+2)*invb4} \equiv \var{a5} \pmod{\var{n}}$ .

Finding the inverse:

Use the Euclidean Algorithm to find $s$ and $t$ such that $\simplify{{b4}s+{n}t = gcd({b4},{n}) = 1}$ .

It follows that $s \bmod n$ is the inverse of $\var{b4}$ in $\mathbb{Z}_{\var{n}}$ .

In this case one such application gives:

{describegcd(n,b4)}

Using this we obtain:

$\simplify[]{{extendedgcd2(n,b4)}*{b4}+{extendedgcd1(n,b4)}*{n}=1}$

Hence the inverse of $\var{b4}$ in $\mathbb{Z}_{\var{n}}$ is $\simplify{{extendedgcd2(n,b4)}} \equiv \var{invb4} \pmod{\var{n}}$ .

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Solve linear equations in the rationals and Z_n

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Bill Foster 6 years, 2 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

Parts

Gap-fill

a)

b)

Finding the inverse:

c)

Finding the inverse:

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Solve linear equations in the rationals and Z_n

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Bill Foster 6 years, 2 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

a)

b)

Finding the inverse:

c)

Finding the inverse:

Generated value: `integer`